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1

$begingroup$

The conditions for the problem I am currently tackling is as follows:

Let $T, M > 0$. Define :

$X_T := $

where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$

Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.

For $a in mathbbR$ and $u in C^1 ([0,T])$ define:

$psi(u)(t) = a + int^t_0 f(u(s)) textds$

Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.

The question for this problem is as follows:

Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.

I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.

I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?

Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.

functional-analysis functions norm normed-spaces upper-lower-bounds

share|cite|improve this question

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
  $endgroup$
  – David Hughes
  Mar 7 at 16:48
  

  
  
  
  

add a comment | 

1

$begingroup$

The conditions for the problem I am currently tackling is as follows:

Let $T, M > 0$. Define :

$X_T := $

where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$

Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.

For $a in mathbbR$ and $u in C^1 ([0,T])$ define:

$psi(u)(t) = a + int^t_0 f(u(s)) textds$

Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.

The question for this problem is as follows:

Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.

I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.

I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?

Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.

functional-analysis functions norm normed-spaces upper-lower-bounds

share|cite|improve this question

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I feel foolish now. The key is probably the fact that the norm of $u$ is always less than or equal to $2M$. I'm working on it now...
  $endgroup$
  – David Hughes
  Mar 7 at 16:48
  

  
  
  
  

add a comment | 

$begingroup$

The conditions for the problem I am currently tackling is as follows:

Let $T, M > 0$. Define :

$X_T := $

where $||u||_T = ^textsup_t in (0,T) [|u(t)| +|u'(t)|]$ is the norm of $C^1 ([0,T])$

Suppose that $f : mathbbR rightarrow mathbbR$ belongs to $C^1 (mathbbR)$ and that $f(0) = 0$.

For $a in mathbbR$ and $u in C^1 ([0,T])$ define:

$psi(u)(t) = a + int^t_0 f(u(s)) textds$

Small note, not sure if this is relevant, but I have previously shown that $X_T$ as defined above is a complete metric space under the above norm.

The question for this problem is as follows:

Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $psi$ defines a map from $X_T$ to itself.

I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u in X_T$ but I can't extend this to a general value that works for all $u$.

I would like to be able to find an upper bound for $^textsup_t in (0,T) |f(u(t))|$ over all $u in X_T$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $mathbbR$. We would need $f$ to be bounded on $mathbbR$ for this to work right?

Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.

functional-analysis functions norm normed-spaces upper-lower-bounds

share|cite|improve this question

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

$endgroup$

share|cite|improve this question

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

share|cite|improve this question

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

asked Mar 7 at 16:43

David HughesDavid Hughes

1607

1 Answer
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0

$begingroup$

It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$

share|cite|improve this answer

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

$endgroup$

add a comment | 

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0

$begingroup$

It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$

share|cite|improve this answer

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

$endgroup$

add a comment | 

0

$begingroup$

It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$

share|cite|improve this answer

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

$endgroup$

add a comment | 

$begingroup$

It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| leq 1$ on some $(−2M,2M)$

share|cite|improve this answer

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

$endgroup$

share|cite|improve this answer

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

answered Mar 21 at 18:42

David HughesDavid Hughes

1607

answered Mar 21 at 18:42

David HughesDavid Hughes

1607